Chapter 5: Factorization of Algebraic Expressions Exercise – 5.1

Question: 1

Find the value

x³ + x − 3x² − 3

Solution:

Taking x common in x³ + x

= x(x² + 1) − 3x² − 3

Taking - 3 common in − 3x² − 3

= x(x² + 1) − 3(x² + 1)

Now, we take (x² + 1) common

= (x² + 1) (x - 3)

 ∴ x³ + x − 3y² − 3 = (x² + 1)(x - 3)

Question: 2

Find the value

a(a + b)³ − 3a²b(a + b)

Solution:

Taking (a + b) common in the two terms

= (a + b){a(a + b)² - 3a²b}

Now, using (a + b)²= a² + b² + 2ab

= (a + b){a(a² + b² + 2ab) − 3a²b}

= (a + b){a³ + ab² + 2a²b − 3a²b}

= (a + b){a³ + ab² − a²b}

= (a + b)p{a² + b² − ab}

= p(a + b)(a² + b² − ab)

∴ a(a + b)³ − 3a²b(a + b)

= a(a + b)(a² + b² − ab)

Question: 3

Find the value

x(x³ − y³) + 3xy(x − y)

Solution:

Elaborating x³ − y³ using the identity  

x³ − y³ = (x − y)(x² + xy + y²)

= x(x − y)(x² + xy + y²) + 3xy(x − y)

Taking common x(x - y) in both the terms

= x(x − y)(x² + xy + y² + 3y)

∴ x(x³ − y³) + 3xy(x − y)

= x(x − y)(x² + xy + y² + 3y)

Question: 4

Find the value

a²x² + (ax² + 1)x + a

Solution:

We multiply x(ax² + 1) = ax³ + x

= a²x² + ax³ + x + a

Taking common ax² in (a²x² + ax³) and 1 in (x + a)

= ax²(a + x) + 1(x + a)

=ax²(a + x) + 1(a + x)

Taking (a + x) common in both the terms

= (a + x)(ax² + 1)

∴ a²x² + (ax² + 1)x + a = (a + x)(ax² + 1)

Question: 5

Find the value

x² + y − xy − x

Solution:

On rearranging

x² − xy − x + y

Taking x common in the (x² − xy) and -1 in(-x + y)

= x(x - y) - 1 (x - y)

Taking (x - y) common in the terms

= (x - y)(x - 1)

∴ x² + y − xy − x = (x - y)(x - 1)

Question: 6

Find the value

x³ − 2x²b + 3xy² − 6y³

Solution:

Taking x² common in (x³ − 2x²y) and +3y² common in (3xy² − 6y³)

= x²(x − 2y) + 3y²(x − 2y)

Taking (x - 2y) common in the terms

= (x − 2y)(x² + 3y²)

∴ x³ − 2x²y + 3xy² − 6y³ = (x − 2y)(x² + 3y²)

Question: 7

Find the value

6ab − b² + 12ac − 2bc

Solution:

Taking b common in (6ab − b²) and 2c in (12ac - 2bc)

= b(6a - b) + 2c (6a - b)

Taking (6a - b) common in the terms

= (6a - b)(b + 2c)

∴ 6ab − b² + 12ac − 2bc = (6a - b)(b + 2c)

Question: 8

Find the value

[x² + 1/x²] − 4[x + 1/x] + 6

Solution:

= x² + 1/x² − 4x − 4/x + 4 + 2

= x² + 1/x² + 4 + 2 − 4x − 4x

= (x²) + (1/x)² + (−2)² + 2 × x × 1/x + 2 × 1/x × (−2) + 2(−2)x

Using identity

x² + y² + z² + 2xy + 2yz + 2zx = (x + y + z)²

We get,

= [x + 1/x + (−2)]²

= [x + 1/x − 2]²

= [x + 1/x − 2][x + 1/x − 2]

∴ [x² + 1/x²] − 4[x + 1/x] + 6 = [x + 1/x − 2][x + 1/x − 2]

Question: 9

Find the value

x(x - 2)(x - 4) + 4x - 8

Solution:

= x(x  - 2)(x - 4) + 4(x - 2)

Taking (x - 2) common in both the terms

=(x - 2){x(x - 4) + 4}

=(x - 2){x² − 4x + 4}

Now splitting the middle term of x² − 4x + 4

= (x - 2){x² − 2x − 2x + 4}

= (x - 2){x( x - 2) -2(x - 2)}

= (x - 2){(x - 2)(x - 2)}

= (x - 2)(x - 2)(x - 2)

= (x − 2)³

∴ x(x - 2)(x - 4) + 4x - 8 = (x − 2)³

Question: 10

Find the value

(x + 2)(x² + 25) − 10x² − 20x

Solution:

(x + 2)(x² + 25) - 10x (x + 2)

Taking (x + 2) common in both the terms

= (x + 2)(x² + 25 − 10x)

= (x + 2)(x² − 10x + 25)

Splitting the middle term of (x² − 10x + 25)

= (x + 2)(x² − 5x − 5x + 25)

= (x + 2){x(x - 5)-5 (x - 5)}

= (x + 2)(x - 5)(x - 5)

∴ (x + 2)(x² + 25) − 10x² - 20x = (x + 2)(x - 5)(x - 5)

Question: 11

Find the value

2a²+2√6ab +3b²

Solution:

Using the identity (p + q)² = p² + q² + 2pq

Question: 12

Find the value

(a − b + c)² + (b − c + a)² + 2(a − b + c) × (b − c + a)

Solution:

Let (a - b + c) = x and (b - c + a) = y

= x² + y² + 2xy

Using the identity (a + b)² = a² + b² + 2ab

= (x + y)²

Now, substituting  x and y

(a - b + c + b − c + a)²

Cancelling -b, +b  & + c, -c

= (2a)²

= 4a²

∴ (a − b + c)² + (b − c + a)² + 2(a − b + c) × (b − c + a) = 4a²

Question: 13

Find the value

a² + b² + 2(ab + bc + ca)

Solution:

= a² + b² + 2ab + 2bc + 2ca

Using the  identity (p + q)² = p² + q² + 2pq

We get,

= (a + b)² + 2bc + 2ca

= (a + b)² + 2c(b + a)

Or (a + b)² + 2c(a + b)

Taking (a + b) common

= (a + b)(a + b + 2c)

∴ a² + b² + 2(ab + bc + ca) = (a + b)(a + b + 2c)

Question: 14

Find the value

4(x − y)² − 12(x − y)(x + y) + 9(x + y)²

Solution:

Let(x - y) = x,(x + y) = y

= 4x² − 12xy + 9y²

Splitting the middle term - 12 = – 6 - 6  also 4 × 9 = −6 × − 6

= 4x² − 6xy − 6xy + 9y²

= 2x(2x - 3y) -3y(2x - 3y)

= (2x - 3y)(2x - 3y)

= (2x − 3y)²

Substituting x = x - y & y = x + y

= [2(x − y) − 3(x + y)]² = [2x - 2y - 3x - 3y]²

= (2x - 3x - 2y - 3y)²

= [−x − 5y]²

= [(−1)(x + 5y)]²

= (x + 5y)²    [? (-1)² = 1]

∴ 4(x − y)² − 12(x − y)(x + y) + 9(x + y)² = (x + 5y)²

Question: 15

Find the value

a² − b² + 2bc − c²

Solution:

a² − (b² − 2bc + c²)

Using the identity (a − b)² = a² + b² − 2ab

= a² − (b − c)²

Using the identity a² − b² = (a + b)(a − b)

= (a + b - c)(a - (b - c))

= (a + b - c)(a - b + c)

∴ a² − b² + 2bc − c² = (a + b - c)(a - b + c)

Question: 16

Find the value

a² + 2ab + b² − c²

Solution:

Using the identity (p + q)² = p² + q² + 2pq

= (a + b)² − c²

Using the identity p² − q² = (p + q)(p − q)

= (a + b + c)(a + b - c)

∴ a² + 2ab + b² − c² = (a + b + c)(a + b - c)

Question: 17

Find the value

a² + 2ab + b² − c²

Solution:

Using the identity (p + q)² = p² + q² + 2pq

= (a + b)² − c²

Using the identity p² − q² = (p + q)(p − q)

= (a + b + c)(a + b - c)

∴  a² + 2ab + b² − c² = (a + b + c)(a + b - c)

Question: 18

Find the value

xy⁹ − yx⁹

Solution:

= xy(y⁸ − x⁸)

= xy((y⁴)² − (x⁴)²)

Using the identity p² − q² = (p + q)(p - q)

= xy(y⁴ + x⁴)(y⁴ − x⁴)

= xy(y⁴ + x⁴)((y²)² − (x²)²)

Using the identity p² − q² = (p + q)(p - q)

= xy(y⁴ + x⁴)(y² + x²)(y² − x²)

= xy(y⁴ + x⁴)(y² + x²)(y + x)(y − x)

= xy(x⁴ + y⁴)(x² + y²)(x + y)(−1)(x − y)

∴ (y − x) = −1(x − y)

= −xy(x⁴ + y⁴)(x² + y²)(x + y)(x − y)

∴ xy⁹ − yx⁹ = −xy(x⁴ + y⁴)(x² + y²)(x + y)(x − y)

Question: 19

Find the value

x⁴ + x²y² + y⁴

Solution:

Adding x²y² and subtracting x²y² to the given equation

= x⁴ + x²y² + y⁴ + x²y² − x²y²

= x⁴ + 2x²y² + y⁴ − x²y²

= (x²)² + 2 × x² × y² + (y²)² − (xy)²

Using the identity (p + q)² = p² + q² + 2pq

= (x² + y²)² − (xy)²

Using the identity  p² − q² = (p + q)(p - q)

= (x² + y² + xy)(x² + y² − xy)

∴  x⁴ + x²y² + y⁴ = (x² + y² + xy)(x² + y² − xy)